Calculus 8th Edition

$\frac{1}{\pi}(1-e^{-2\pi})$
$\int_0^2\frac{dx}{e^{\pi x}}$ $=\int_0^2 e^{-\pi x}dx$ Let $u=-\pi x$. Then $du=-\pi dx$, and $\frac{1}{-\pi}du=dx$. $=\int_{-\pi*0}^{-\pi*2}e^u* \frac{1}{-\pi} du$ $=\frac{1}{-\pi}\int_0^{-2\pi} e^u du$ $=-\frac{1}{\pi}(e^{-2\pi}-e^0)$ $=-\frac{1}{\pi}(e^{-2\pi}-1)$ $=\frac{1}{\pi}(1-e^{-2\pi})$